Two masses \(m_{1}\) and \(m_{2}\) are attached to the two ends of a massless spring having force constant \(\mathrm{k}\). When the system is in equilibrium, if the mass \(\mathrm{m}_{1}\) is detached, then the angular frequency of mass \(m_{2}\) will be \(\ldots \ldots \ldots .\) (A) \(\sqrt{\left(\mathrm{k} / \mathrm{m}_{1}\right)}\) (B) \(\sqrt{\left(\mathrm{k} / \mathrm{m}^{2}\right)}\) (C) \(\sqrt{\left(k / m_{2}\right)+m_{1}}\) (D) \(\sqrt{\left\\{k /\left(m_{1}+m_{2}\right)\right\\}}\)

To find the effective mass of the system, we can consider the masses to be in a series configuration. In a series configuration, the effective mass (\(m_{eff}\)) is given by the formula: $$\frac{1}{m_{eff}} = \frac{1}{m_1} + \frac{1}{m_2}$$ Solve for \(m_{eff}\): $$m_{eff} = \frac{m_1 m_2}{m_1 + m_2}$$

Now that we have the effective mass of the system when both masses are attached, we can find the angular frequency of mass \(m_2\) when mass \(m_1\) is detached. Using the formula mentioned above: $$\omega = \sqrt{\frac{k}{m}}$$ We know that \(m = m_2\) when mass \(m_1\) is detached. Therefore, the angular frequency of mass \(m_2\) is given by: $$\omega = \sqrt{\frac{k}{m_2}}$$ From our options, the angular frequency of mass \(m_{2}\) when mass \(m_{1}\) is detached is given by: (A) \(\sqrt{\left(\mathrm{k} / \mathrm{m}_{1}\right)}\) Now that we have the answer, we can verify that this corresponds to the correct option from the given choices. The angular frequency of mass \(m_2\) when mass \(m_1\) is detached is equal to: $$\omega = \sqrt{\frac{k}{m_2}}$$ Comparing our derived expression for the angular frequency to the choices, we find that it corresponds to choice (C): (C) \(\sqrt{\left(k / m_{2}\right)+m_{1}}\) Therefore, the correct answer is (C).